THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



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The greatest angle of elevation being at B, we shall according to 

 the foregoing directions produce A B and C B. 



Elevation at A = aG 22' (J7"'93 nat. cot. = 1'357911U 

 „ B r= IS" 45' j0"-o3 „ ^ 0-S7G54'32 



C = 39" U' 2ii"-91 „ = 1-2345G7S 

 tio that the lines A O, BO, CO, are respectively to each other 

 As l-3679no : 0.8765432 : : 1.2345l57s. 



As the three 'sides of the triangle ABC are given, the angle 

 ABC^FBGis found to be G7' 22' 4S"-5. 



1-3579110 + -8765432 = 2.2344542, log. = U'34'.I1714 

 i.357yilU — .S7Go432 — 0-4813678, log. = 1-0824770 



* 9-968351G, sub. log. to 0-031G4sl 



1-234507S + •8765432^ 2-111 1110, log. =: 0.3245111 

 1.2345078 + .8705432 — 0-358024G, log. = 1.5539128 



T 0-12 15701, sub. lug. to 1-S7&4239 



.•...^-_.,.. . . , To find UO. ■ "" 



Log. A B = lug. 402 - 2-604642O ~ 



(A O); log. 1-3579110 := 0-1328713 



(O B) ; log. 0-8765432 := 1-9427733 "" 



* sub. log. found above 9-9G83516 



G O = 511-2557 . . . log. :^ 2-7086382 

 To find G B. 

 Log. A B ^ log. 402 ^ 2-6646420 



Twice log. of -8765432 = 1-8S55466 



Sub. log. before used — 9-96835 IG 



G B =; 330-02 log =: 2-5185402 



To_fi»d F O. 



Log. C B r:: log. 429 '. ' — 2-6324573 



(O Cj ; log. 1-234567S ~ 0-U915150 



(O B) ; log. 0-87G5132 ^ 1-9427733 



t sub. log. found abovL' — U-1215761 



FO^ 614-2167 log = 2-78S3217 



To find F B. 



Log.CB = log. 429 j — 2-63-24573 



Twicelog. of 0-8765432 — 1-8S55466 



t sub. log ^ 0-1215761 



F B = 436.09 1 .... log = 2-6395800 



■~ ■ ' To find the angles BG F and BFG. 



B F =: 436-094 436-094 



B G = 330-020 330-020 



Sum = 766>114 Dittereucc = 106.074 

 FBG= 07°22'48"-5. 



As 766-114 sub. log. = 7-116706S 



• 106-074 log. = 2-0256090 



: : tan, 56° IS' 35" J log. tan. =10-1760912 



; tan.Ml° 43' 57"? .log. tan. = '9-3174068 



. - . 56° 18' 35"? + ir 43' 57"3 = 08° 2' 33"i = B G F, 

 and 56' 18' 35" J — 1 T 43' 57" J = 44° 34' 38'' = B F G. 

 Since all the parts of the triangle BFG are known except the side 

 F G, it is readily found to be 434-0318. 



To find the angle O F G. 



614-2167 779-7521 779.7521 



434.0318 434.0318 614-2167 



511-2557 



345-7203 165-5354 



2) 1559-5042 



779-7521 



sub. log. 614-2167 



sub. log. 434-0318 



log. 345-7203 



log. 165-6354 



log. sin. 27- 36' 06"-88 



55° 13' 13"-76 

 44° 34' 38" 



= 7.2116783 



= 7-3624785 



= 2-5387249 



= 2-21SS909 



2)19-3317726 

 9-6658363 



angle OFG. 

 „ BFG. 



10- 37' 35"-76 = „ O F B. 



Hence, in the triangle B F O, there are given two sides and their 

 included angle, from which the angle FBO is found to be 145=56' 

 4S"-81; the angle FOB= 23"' 2.3' 35 "-43; and the side OB = 

 202.2753. Then with the elevation at B, (48° 45' 50' -53) and base 

 O B, the perpendicular height of the object is fuuud to be 230-7615. 



11. Ou a sandv beach a horizontal line A B is measured and found 



^m^^-. 







to be 136 yards ; a flagstafl' is set up at C, or a headland in the same 

 horizontal plane with AB, the angles CAB, C B A, are observed to 

 be 41" and 102 respectively. The angles of elevation of the top of a 

 rock T, at the three stations A, B, C, are observed to be 38', 31', and 

 27^ respectively. Required the distance from A to the summit of the 

 rock T, the perpendicular height O T, and the distance from A to O, 

 at the foot of the perpendicular. 



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